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kogti [31]
2 years ago
5

The sum of 2 numbers is 84 if one number is 4.less than the other find the numbers​

Mathematics
2 answers:
faust18 [17]2 years ago
7 0
Hope this helps !! Did it step by step

artcher [175]2 years ago
3 0

Answer:

one number= x = 44

the other number= (x-4)= 40

Step-by-step explanation:

one number= x

the other number= (x-4)

x + (x-4) = 84

2x= 84+4

2x= 88

x= 88/2= 44

one number= x = 44

the other number= (x-4)= 40

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Write an equation of the line that passes through the points.<br><br> (2, 5), (0, 5)<br><br> Y =
vodomira [7]
Right off the bat we know that the y-intercept is going to be 5 because of the second co-ordinate.

Now for the slope, because the y co-ordinate doesn't change, it has no slope so the equation is simply: y=5

Hope this helps :)
7 0
3 years ago
A system of equations is shown below. y = –x – 3 5x + 2y = 9 What is the solution to this system?
erastovalidia [21]

Answer:

x=5\ and\ y=-8

The solution of the system can be given as (5,-8).

Step-by-step explanation:

Given system of equations:

y=-x-3

5x+2y=9

To solve the given system of equations:

Solution:

We will use substitution to solve the given system.

We will substitute the y values in terms of x from the equation y=-x-3 into the other equation to solve for x

Substituting y=-x-3 into 5x+2y=9.

We have:

5x+2(-x-3)=9

<em>Using distribution.</em>

5x-2x-6=9

<em>Combining like terms.</em>

3x-6=9

Adding 6 both sides.

3x-6+6=9+6

3x=15

Dividing both sides by 3.

\frac{3x}{3}=\frac{15}{3}

∴ x=5

<em>Plugging in x=5 into y=-x-3.</em>

We have,

y=-5-3

∴ y=-8

The solution of the system can be given as (5,-8).

4 0
3 years ago
A slitter assembly contains 48 blades. Five blades are selected at random and evaluated each day of sharpness. If any dull blade
Alex73 [517]

Answer:

Part a

The probability that assembly is replaced the first day is 0.7069.

Part b

The probability that assembly is replaced no replaced until the third day of evaluation is 0.0607.

Part c

The probability that the assembly is not replaced until the third day of evaluation is 0.2811.

Step-by-step explanation:

Hypergeometric Distribution: A random variable x that represents number of success of the n trails without replacement and M represents number of success of the N trails without replacement is termed as the hypergeometric distribution. Moreover, it consists of fixed number of trails and also the two possible outcomes for each trail.

It occurs when there is finite population and samples are taken without replacement.

The probability distribution of the hyper geometric is,

P(x,N,n,M)=\frac{(\limits^M_x)(\imits^{N-M}_{n-x})}{(\limits^N_n)}

Here x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.

Probability that at least one of the trail is succeed is,

P(x\geq1)=1-P(x

(a)

Compute the probability that the assembly is replaced the first day.

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48.

Number of blades selected at random from the assembly  n= 5

Number of blades in an assembly dull is M  = 10.

The probability mass function is,

P(X=x)=\frac{[\limits^M_x][\limits^{N-M}_{n-x}]}{[\limits^N_n]};x=0,1,2,...,n\\\\=\frac{[\limits^{10}_x][\limits^{48-10}_{5-x}]}{[\limits^{48}_5]}

The probability that assembly is replaced the first day means the probability that at least one blade is dull is,

P(x\geq 1)=1- P(x

(b)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly  N = 48

Number of blades selected at random from the assembly  N = 5

Number of blades in an assembly dull is  M = 10

From the information,

The probability that assembly is replaced (P)  is 0.7069.

The probability that assembly is not replaced is (Q)  is,

q=1-p\\= 1-0.7069= 0.2931

The geometric probability mass function is,

P(X = x)= q^{x-1} p; x =1,2,....=(0.2931)^{x-1}(0.7069)

The probability that assembly is replaced no replaced until the third day of evaluation is,

P(X = 3)=(0.2931)^{3-1}(0.7069)\\=(0.2931)^2(0.7069)= 0.0607

(c)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly   N = 48

Number of blades selected at random from the assembly  n = 5

Suppose that on the first day of the evaluation two of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M  = 2.

P(x=0)=\frac{(\limits^2_0)(\limits^{48-2}_{5-0})}{\limits^{48}_5}\\\\=\frac{(\limits^{46}_5)}{(\limits^{48}_5)}\\\\= 0.8005

Suppose that on the second day of the evaluation six of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M  = 6.

P(x=0)=\frac{(\limits^6_0)(\limits^{48-6}_{5-0})}{(\limits^{48}_5)}\\\\=\frac{(\limits^{42}_5}{(\limits^{48}_5)}\\\\= 0.4968

Suppose that on the third day of the evaluation ten of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M

= 10.

P(x\geq 1)=1- P(x

 

The probability that the assembly is not replaced until the third day of evaluation is,

P(The assembly is not replaced until the third day)=P(The assembly is not replaced first day) x P(The assembly is not replaced second day) x P(The assembly is replaced third day)

=(0.8005)(0.4968)(0.7069)= 0.2811

5 0
3 years ago
Which expression is equivalent to -0.55 – 0.47a + a?
Leto [7]

Answer:

-0.55 + 0.53a

Step-by-step explanation:

              -0.55 – 0.47a + a

 To find an expression equivalent to this, we must simplify the equation to a reasonable extent;

         -0.55 – 0.47a + a

      = -0.55 + 0.53a

      = 0.53a - 0.55

The expression equivalent to the given one is -0.55 + 0.53a  or 0.53a - 0.55

8 0
2 years ago
3 - x3 &gt; -4 help pls
Kamila [148]

Answer:

x<1.912931

Step-by-step explanation:

6 0
3 years ago
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